Sum of areas of traingles APB and CEPD

Question

In triangle $ABC$ points $D$ and $E$ lie on the sides $BC$ and $AC$ respectively, such that $\frac {BD} {DC} = \frac {CE} {EA} = 2.$

$AD \cap BE = P$ (a point). Given that the area of $BPD$ is $252$ more than the area of $APE$, find the sum of the areas of $\triangle APB$ and $CEPD$.

I am looking for a solution without using Menelaus.

Answer 1

Let $DC=3x$ and $AE=3y$. Hence, $BD=6x$ and $EC=6y$.

Now, let $G\in EC$ such that $DG||BE$.

Hence, $EG:GC=2:1$, which gives $EG=4y$ and $AP:PD=AE:EG=3y:4y=3:4$.

Let $F\in DC$,such that $EF||AD$.

Thus, $DF:FC=1:2$, which gives $DF=x$.

Hence, $BP:PE=BD:DF=6x:x=6:1$.

Let $PE=t$ and $AP=3z$.

Hence, $\frac{S_{\Delta BPD}}{S_{\Delta APE}}=\frac{6t\cdot4z}{t\cdot3z}=8$, which says $S_{\Delta APE}=36$ and $S_{\Delta BPD}=288$.

Thus, $S_{\Delta APB}=6\cdot36=216$, $S_{\Delta ADC}=\frac{1}{2}(216+288)=252$

and from here $S_{CEPD}=252-36=216$, which gives the answer: $432$.